3B Wave Motion II
5
Chapter 5 Nature of Waves
Nature of Waves
Practice 5.1 (p. 7)
1
B
2
D
3
(a) Transverse wave
11
Speed = f  = 5  0.2 = 1 m s1
1 1
(ii) Period = = = 0.2 s
f 5
(a) (i)
(b) A heavier string (length unchanged) has
greater mass per unit length. Therefore,
(b) Longitudinal wave
4
(a) Water wave and EM wave
(b) Sound wave
5
12
For case I, energy is transferred to the cork
directly from the stone. For case II, energy is
the wave speed decreases.
distance travelled
(a) Wave speed =
time taken
5
=
3
transferred to the cork through water waves.
6
= 1.6667
An object floating on water vibrates as a wave
= 1.67 m s1
passes it. The energy of the object comes
(b) By v = f,
from what starts the wave, e.g. a stone.
Wavelength =
Practice 5.2 (p. 21)
1
C
2
C
3
A
4
D
5
C
6
C
7
B
8
(a) C
13
9
10
12
=4m
3
12
(b) Wave speed =
= 6 m s1
2
(a) Wavelength =
(c)
By v = f,
frequency =
v 6
= = 1.5 Hz
λ 4
Practice 5.3 (p. 29)
(b) A
(c)
v 1.6667
=
= 0.833 m
f
2
B
4
= 0.16 m
25
distance travelled
(b) Speed =
time taken
4
=
= 0.4 m s1
10
v 0 .4
(c) Frequency = =
= 2.5 Hz
λ 0.16
1
B
2
A
3
(a) Wavelength = 20 cm
λ 0 .2
(b) Speed = f = =
= 2 m s1
T 0 .1
4
(a) Amplitude = 50 cm
(a) Wavelength =
(b) Time taken = 2  2 = 4 s
5
From the graph, the period of the wave is 4 s.
By v = f,
wavelength =
W and Y are momentarily at rest.
v
= vT = 10  4 = 40 m
f
X is moving upwards.
Z is moving downwards.
New Senior Secondary Physics at Work
1
 Oxford University Press 2009
3B Wave Motion II
Chapter 5 Nature of Waves
(b) (i)
6
Particles X and Z are in phase.
(ii) Particles W and Y are in antiphase.
(iii) Particle W is on a wave crest at
t = 0.6 s.
8
(a) (i)
Wavelength = 6.4 cm
(ii) Since particle P undergoes the
smallest number of oscillation at
the instances shown, P should have
1
oscillated for period from t = 0
4
to t = 0.5 s.
Frequency =
0.25
= 0.5 Hz
0 .5
(iii) Speed = f
= 0.5  0.064
= 0.032 m s1
(b) The wave is travelling towards the right.
Revision exercise 5
Multiple-choice (p. 34)
1
C
2
A
3
C
4
D
From the displacement–distance graph,
7
(a) (i)
wavelength of the wave is 0.5 m.
Amplitude = 15 cm
By v = f,
(ii) Wavelength = 2 m
frequency =
(iii) Particle W has at least oscillated for
0.5 period from t = 0 to t = 0.2 s.
0 .5
Minimum frequency =
0 .2
5
A
6
C
v 20
=
= 40 Hz
λ 0.5
By v = f,
= 2.5 Hz
v 12
=
= 4 km
3
f
1
A and B are 1  apart. Therefore, when A is
4
wavelength =
(iv) Minimum wave speed
= minimum frequency  
= 2.5  2 = 5 m s1
on a crest, B is at its equilibrium position.
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2
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3B Wave Motion II
7
Chapter 5 Nature of Waves
C
(b)
The maximum wavelength is 0.4 m.
Maximum speed
= frequency  maximum wavelength
= 10  0.4
= 4 m s1
8
C
(Correct wavelength)
(1A)
Period = 16 s
(Correct amplitude)
(1A)
Wavelength = 40 cm
λ 0.4
Speed = f = =
= 0.025 m s–1
T 16
4
Distance travelled
= vt = 0.025  4 = 0.1 m = 10 cm
9
D
10
(HKALE 2003 Paper II Q14)
11
D
12
(HKCEE 2005 Paper II Q15)
13
(HKCEE 2005 Paper II Q34)
14
(HKCEE 2005 Paper II Q35)
15
D
16
(HKCEE 2006 Paper II Q16)
(Correct waveform after 1 s)
(1A)
distance travelled
(a) Speed of pulse =
(1M)
time taken
5
=
1
= 5 m s1
(b) By v = f,
(c)
5
(a) Transverse pulse is generated.
The wavelength decreases.
(1A)
(a) (i)
(1M)
(b) (i)
(a) The wavelength is 0.2 m.
downwards.
(c)
(2  1A)
Particle Q will return to the equilibrium
position.
(1M)
New Senior Secondary Physics at Work
(2  1A)
(1A)
Particles O and S are moving
v 0.01
=
= 0.25 Hz
λ 0.04
(2  1A)
(iii) Particle Q is momentarily at rest.
(1A)
(a) By v = f,
(1A)
(ii) Particles P and T are moving
(2  1A)
Particle Q is moving upwards.
frequency =
(1A)
Particles R and S are moving
upwards.
(1A)
(b) Particles P and R are momentarily at
downwards.
Amplitude = 2 cm
= 0.25 Hz
(1A)
rest.
(1A)
(ii) Wavelength = 2  4 = 8 cm (1A)
16
(iii) Period =
=4s
(1A)
4
1
(iv) Frequency =
(1M)
T
1
=
4
(1A)
= 2 m s1
3
(1A)
spring with the same length.
(b) Speed of the pulse
distance travelled
=
time taken
2
=
1
2
(1M)
v 5
wavelength = = = 1.25 m
f 4
(d) Stretch the spring more / use a lighter
Conventional (p. 36)
1
(1A)
6
(a) A is moving downwards.
(1A)
(1A)
B and C are moving upwards. (2  1A)
(1A)
3
 Oxford University Press 2009
3B Wave Motion II
Chapter 5 Nature of Waves
(b) The greatest displacement
(c)
(d)
= amplitude of the wave
(1A)
= 1 cm
(1A)
Wavelength = 4 cm
(1A)
(d)
(Label X, Y and Z correctly)
(3  1A)
(e)
(Correct waveform)
(1A)
(Correct positions of A, B and C) (1A)
7
(a) (i)
Amplitude = 2 cm
(1A)
(ii) Wavelength = 5 m
(1A)
(b) Point B is 1 cm from the equilibrium
position
(c)
(i)
(1A)
Frequency
5
=
2
= 2.5 Hz
(1A)
Speed
=f
(1M)
= 2.5  5
= 12.5 m s1
(ii) Time required
distance travelled
=
speed
25
=
12.5
=2s
(1A)
8
(Correct axes)
(1A)
(Correct curve)
(1A)
(Correct period and amplitude)
(1A)
When the frequency of the ‘wave’ produced
by the transverse wave model increases, wave
(1M)
speed increases
(1A)
while wavelength keeps constant.
(1A)
For a real wave on a string, when its
frequency increases, its wavelength decreases
(1A)
(1A)
9
while the wave speed keeps constant.
(1A)
(a) (i)
(1A)
Amplitude = 4 cm
(ii) Wavelength = 4.8 cm
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4
(1A)
 Oxford University Press 2009
3B Wave Motion II
(b) (i)
Chapter 5 Nature of Waves
Period =
1
f
=
1
2
(1M)
= 0.5 s
(1A)
(ii)
11
(Correct labelled axes)
(1A)
(Correct amplitude)
(1A)
(Correct wavelength)
(1A)
(Correct waveform)
(1A)
(a) (i)
10
Frequency =
5
= 2 Hz
12
(ii) Wavelength =
8
(‘A’ correctly marked.)
= 1.5 m
(1A)
(1M)
(1A)
(1M)
(1A)
(iii) Speed = f
(c)
(b) (i)
= 2  1.5
(1M)
1
(1A)
=3ms
15  13
Amplitude =
cm
2
= 1 cm
(1M)
(1A)
(ii) The waves carry less energy at Q
than at P.
12
(HKCEE 2002 Paper I Q4)
13
(a) (i)
(1A)
incorrect
The amplitude of a wave is the
maximum displacement of a
10
(Correct labelled axes)
(1A)
(Correct amplitude)
(1A)
(Correct period)
(1A)
(Correct waveform)
(1A)
particle in the wave from its
equilibrium position.
(ii) correct
(1A)
(1A)
(iii) incorrect
The distance travelled by the wave
From the figure, the period of the wave is 2 s.
in the periodic time.
The wave has travelled a distance of one
(iv) correct
wavelength from t = 0 to t = 2 s. Therefore,
(1A)
(1A)
the displacementdistance graph of the wave
is as follows.
New Senior Secondary Physics at Work
5
 Oxford University Press 2009
3B Wave Motion II
Chapter 5 Nature of Waves
(b) Any two of the following:
(2  1A)
(3) Period =
The velocity of the wave is in the
1
1
=
= 2.0 s (1A)
f 0.5
direction of propagation while the
velocity of a piece of rope is at right
Physics in articles (p. 40)
angles to this.
(a) It is a transverse wave
The velocity of the wave is constant
because each person’s motion is
while the velocity of a piece of rope
perpendicular to the direction of propagation.
(1A)
varies (with time or displacement).
(b) The wavelength of a typical Mexican wave is
The velocity of the wave is non-zero
(c)
while the average velocity of a piece of
9 m.
(1A)
rope is zero.
By v = f,
(1M)
Energy spreads out over larger
frequency =
circumference / energy is continuously
lost from wave.
14
(1A)
(a) (i)
(c)
(1A)
Transverse wave
(ii) EM wave
v 12
=
= 1.33 Hz
λ 9
Similarity: Both waves on a string and
Mexican waves transfer a disturbance from
(1A)
one place to another.
(1A)
while Mexican waves do not.
(Arrow correctly drawn)
(1A)
(iii) (Q correctly labelled)
(1A)
(iv) Wavelength
(1A)
(v) (R correctly labeled)
(1A)
=
(1A)
(1A)
(ii) 0.4 m
(vi) (1) Speed
distance travelled
=
time taken
(1A)
Difference: Waves on a string transfer energy
(b)
(i)
(1A)
(1M)
0 .1
0.20
= 0.5 m s1
(2) By v = f,
(1A)
(1M)
v
frequency =
λ
0 .5
=
1 .0
= 0.5 Hz
New Senior Secondary Physics at Work
(1A)
6
 Oxford University Press 2009